Login

Welcome, Guest. Please login or register.

March 30, 2024, 02:14:06 am

Author Topic: Turning point of parabola  (Read 1173 times)  Share 

0 Members and 1 Guest are viewing this topic.

#1procrastinator

  • Guest
Turning point of parabola
« on: August 08, 2011, 01:26:47 pm »
0
Why is the x-coordinate of a quadratic graph given by -b/2a? I saw the derivation but I don't understand the relationship between the two constants and why it would give the middle point of the graph

I'm curious about the y-coordinate too, but I can't remember it haha

b^3

  • Honorary Moderator
  • ATAR Notes Legend
  • *******
  • Posts: 3529
  • Overloading, just don't do it.
  • Respect: +631
  • School: Western Suburbs Area
  • School Grad Year: 2011
Re: Turning point of parabola
« Reply #1 on: August 08, 2011, 02:37:11 pm »
+3
ok so you know the two x-intercepts are given by ,i.e. and . Each of the x-intercepts are the same distance away (along the x-axis) from the turning point. So if you add then together and divide by 2, you should get the x coordinate of the turning point.



The reason it gives the middle point is because parabolas are symmetrical on the y-axis around the turning point, i.e. the left and right sides of the curve are the same, but flipped.

as for the y coordinate
your equation of your parabola is y=ax2+bx+c
so sub it in

which just gives c.

Another way of looking at the y-coordinate is that for it to be a y-intercept, x must be 0. So that means that no matter what the coeffcients of x^2 and x are, they will equal 0 leaving just c.

Or look at it this way. You have your graph of y=ax2+bx=x(ax+b)
so there will be an intercept at x and -b/a
if you shift the graph up c units, there will be a y-intecept (this is previously the point (0,0)) at (0,c)
2012-2016: Aerospace Engineering/Science (Double Major in Applied Mathematics - Monash Uni)
TI-NSPIRE GUIDES: METH, SPESH

Co-Authored AtarNotes' Maths Study Guides


I'm starting to get too old for this... May be on here or irc from time to time.

#1procrastinator

  • Guest
Re: Turning point of parabola
« Reply #2 on: August 09, 2011, 08:47:17 pm »
0
That derivation makes a lot more sense than the one my book gave, which was based on completing the square. Bit of a stupid question, but why does it work for parabolas with no x-intercept? (aside from the reason that the negative square roots cancel out anyway)

Thank you

b^3

  • Honorary Moderator
  • ATAR Notes Legend
  • *******
  • Posts: 3529
  • Overloading, just don't do it.
  • Respect: +631
  • School: Western Suburbs Area
  • School Grad Year: 2011
Re: Turning point of parabola
« Reply #3 on: August 09, 2011, 09:15:28 pm »
0
Just look at it the same way but start with a graph that has x-intercepts, then shift it up so many units so that it doesn't have x-intercepts. As you shift it up (or down) only, the turning point stays at the same value of x.
2012-2016: Aerospace Engineering/Science (Double Major in Applied Mathematics - Monash Uni)
TI-NSPIRE GUIDES: METH, SPESH

Co-Authored AtarNotes' Maths Study Guides


I'm starting to get too old for this... May be on here or irc from time to time.

Graphite

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 395
  • Respect: +6
Re: Turning point of parabola
« Reply #4 on: August 09, 2011, 09:27:12 pm »
+3
You can actually just differentiate and solve for x when dy/dx=0 for that general parabolae function